GCSE Maths Fractions (Higher): check-up and exam-style questions

Method. To divide by a fraction, turn the second fraction upside down and multiply: keep, change, flip.

1. Read the question and decide the plan. Dividing by a fraction is done by multiplying by its reciprocal. First look at the type of numbers we have.
   213 ÷ 134
2. Change the mixed numbers to improper fractions. A mixed number is a whole number plus a fraction. Multiply the whole number by the denominator, add the numerator, and keep the denominator. This lets us calculate with a single fraction.
   213 = (2 × 3 + 1) over 3 = 73
   134 = (1 × 4 + 3) over 4 = 74
3. Keep, change, flip. Dividing by a fraction is the same as multiplying by its reciprocal (the fraction turned upside down). For example, dividing by 1/2 asks "how many halves fit?", which is the same as multiplying by 2.
   73 ÷ 74 = 73 × 47
4. Cancel common factors. Divide a top and a bottom by a number they share, to keep the numbers small.
   73 × 47 = 13 × 41
5. Multiply the numerators, then the denominators. Now it is an ordinary multiplication: top × top and bottom × bottom.
   1 × 4 = 4
   3 × 1 = 3
   43
6. Write the improper fraction as a mixed number. The top is bigger than the bottom, so the answer is more than 1. Divide to find the whole number and the remainder.
   4 ÷ 3 = 1 remainder 1
   43 = 113

Check. Multiply back: 113 × 134 should give 213. 1.3333 × 1.75 = 2.3333 and 213 = 2.3333. ✓

Watch out. Only flip the second fraction (the one you are dividing by), never the first.

Method. Increasing by a fraction: multiply by (1 + the fraction).

1. Find the multiplier. Adding 58 of the amount leaves 1 + 58 of it.
   1 + 58 = 88 + 58 = 138
2. Find one part: divide by the denominator 8. The denominator says to split the amount into 8 equal parts. One part is 18 of the amount.
   72 ÷ 8 = 9
3. Multiply by the numerator 13. The multiplier 138 says we need 13 of those parts.
   9 × 13 = 117
4. Write the answer. The amount has been increased.
   117

Check. A different route: the change is 58 of 72 = 9 × 5 = 45. Then 72 + 45 = 117. ✓

Watch out. The multiplier is 138, not 58. You must add the change to the whole (1).

Method. Simplify the fraction, write its denominator as a product of primes, and see whether any prime other than 2 or 5 appears.

1. Say what the test is. A fraction in its simplest form gives a terminating decimal only if its denominator has no prime factors other than 2 and 5. This is because 10 = 2 × 5, so only denominators made of 2s and 5s can be turned into tenths, hundredths and so on.
   970
2. Check the fraction is in its simplest form. Common factors could change the denominator, so we check first.
   9 and 70 share no factor except 1, so 970 is already simplest.
3. Break the denominator into prime factors. A prime factor is a prime number that divides into it. We divide by the smallest prime each time.
   70 = 2 × 5 × 7
   70 ÷ 2 = 35
   35 ÷ 5 = 7
   7 ÷ 7 = 1
4. Look for primes other than 2 and 5. If there is any other prime factor, the decimal can never end.
   The prime 7 appears, so the decimal recurs.
5. Write the conclusion. Answer the question that was asked.
   970 gives a recurring decimal.

Check. A different route: divide. 9 ÷ 70 = 0.1285714… (a block of digits repeats). ✓

Watch out. Always simplify first. For example 6/15 = 2/5 terminates even though 15 has a factor of 3.

Method. Write each recurring decimal as a fraction, then add the fractions.

1. Write 0.666… as a fraction. Let x = 0.666… Then 10x = 6.666… Subtract: 9x = 6, so x = 6/9.
   0.6 = 69 = 23
2. Write 0.333… as a fraction. Let y = 0.333… Then 10y = 3.333… Subtract: 9y = 3, so y = 3/9.
   0.3 = 39 = 13
3. Add the fractions. The denominators are the same, so add the numerators (2 + 1 = 3) and keep the denominator.
   23 + 13 = 33
4. Simplify. Three thirds is one whole.
   23 + 13 = 1

Check. A different route: 0.666… + 0.333… = 0.999…, and 0.999… = 1. ✓

Watch out. Use the exact fractions, not rounded decimals.

Method. Write a recurring decimal as a fraction using algebra: line up the repeating parts, then subtract to cancel them.

1. Understand the aim. A recurring decimal never ends, so we cannot write it as a fraction by dividing. Instead we use algebra: we make two copies of the number whose repeating tails line up exactly, then subtract so the endless part disappears.
   0.72 = 0.727272…
2. Let x equal the recurring decimal. Using a letter lets us do algebra on the number.
   x = 0.72 = 0.727272…
3. Multiply by 100. 2 digits repeat, so multiplying by 10² = 100 moves exactly one whole block of repeating digits to the left of the point. The tail after the point is then the same as before.
   100x = 72.7272…
4. Subtract the original x. Both numbers have the same infinite tail of digits after the point, so subtracting makes the tail disappear. Remember x means 1x.
   100x − x = 72.7272… − 0.7272…
   99x = 72
5. Solve for x. Divide both sides by the number in front of x.
   x = 7299
6. Simplify the fraction. The highest common factor of 72 and 99 is 9, so divide the top and the bottom by 9.
   7299 = 811

Check. A different route: divide the top by the bottom. 8 ÷ 11 gives 0.72, which matches the original. ✓

Watch out. The number you multiply by (10, 100, 1000…) depends on how many digits repeat and how many do not. Both lines must have the same repeating tail.

Key words. A factor is something that is multiplied. For example 2 and 3 are factors of 6, and (x + 1) is a factor of 3(x + 1). Factorise means rewrite as a product. Cancel means divide the top and the bottom of a fraction by the same factor. An algebraic fraction is just a fraction with letters in it, and it follows exactly the same rules as number fractions.

Method. Simplify an algebraic fraction: (1) factorise the top, (2) factorise the bottom, (3) cancel the factors that match.

1. Write the question and say what is wanted. To simplify means: make the top and bottom as small as possible by cancelling every common factor. The answer must have no common factor left.
   x² − 16x + 4
2. Factorise the top: x² − 16 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
   Spot the pattern. There are two terms, both are perfect squares, and there is a minus between them. This is a difference of two squares.
   The rule is: a² − b² = (a − b)(a + b).
   Find a and b. x² = (x)², so a = x. And 16 = 4², so b = 4.
   Put them into the rule: (x − 4)(x + 4).
   Check by expanding: (x − 4)(x + 4) = x² + 4x − 4x − 16 = x² − 16. ✓ (the middle terms cancel)
   x² − 16 = (x − 4)(x + 4)
3. Factorise the bottom: x + 4 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
   x + 4 is a single bracket already. It cannot be factorised any further.
   x + 4 = (x + 4)
4. Write the fraction again using the factorised top and bottom. Now every factor can be seen and compared.
   (x − 4)(x + 4)(x + 4)
5. Find the factors that are exactly the same on the top and the bottom, and cancel them. Dividing the top and the bottom by the same factor does not change the value of the fraction. A bracket on the top only cancels with a bracket on the bottom that is exactly the same.
   Matching factors: (x + 4).
   (x − 4)(x + 4)(x + 4)
   (the crossed-out factors cancel)
6. Write what is left. This is the answer. Multiply together whatever has not been crossed out.
   (x − 4)

Check. Choose a number for x that does not make any bottom zero. Try x = 2. The original: top = -12, bottom = 6, so the fraction = -2. Our answer at x = 2 = -2. They match. ✓

Watch out. You can only cancel factors (things multiplied), never terms (things added). In (x + 6)/x you cannot cancel the x, because the x is added to 6 on top. That is why we always factorise first.

Key words. The denominator is the bottom of a fraction and the numerator is the top. A common denominator is a bottom that all the fractions share. This works exactly like 1/2 + 1/3 = 3/6 + 2/6 = 5/6, except that the bottoms contain letters.

Method. Add or subtract algebraic fractions: (1) common denominator, (2) rewrite each fraction, (3) combine the tops, (4) expand and collect, (5) simplify if possible.

1. Write the question. We are adding or subtracting algebraic fractions. The answer should be one single fraction.
   x2 + x + 15
2. Find a common denominator. We can only add or subtract pieces that are the same size, so the bottoms must match. The common denominator is the smallest expression that every bottom divides into. Method: factorise each bottom, then take every different factor, using the highest number of copies of it that appear in any one bottom.
   2 = 2
   5 = 5
   Common denominator = 10
3. Rewrite every fraction with the common denominator. Multiplying the top and the bottom by the same thing is like multiplying by 1, so the value of the fraction does not change.
   x2: to turn 2 into 10 we multiply the bottom by 5. We must do the same to the top, so the value does not change:
   5x10
   x + 15: to turn 5 into 10 we multiply the bottom by 2. We must do the same to the top, so the value does not change:
   2(x + 1)10
4. Write one fraction: combine the tops over the common denominator. Now the pieces are the same size, so we add or subtract the tops and keep the bottom.
   5x + 2(x + 1)10
5. Expand each product on the top. Multiply every term in one bracket by every term in the other.
   Expand 5x:
   x × 5 = 5x
   so 5x = 5x
   
   Expand 2(x + 1):
   2 × x = 2x
   2 × 1 = 2
   so 2(x + 1) = 2x + 2
6. Write the top without brackets. There is no minus sign in front of a bracket, so the brackets can simply be removed and the terms written in a row.
   Top = 5x + 2x + 2
7. Collect like terms. Like terms have the same letter part. Add or subtract all the x terms together, and all the plain numbers together.
   Top = 7x + 2
8. Check whether the answer can be simplified. The top might factorise and share a factor with the bottom.
   7x + 2 = (7x + 2)
   (7x + 2)10 = 1/10(7x + 2)

Check. Try x = 2 (it does not make any bottom zero). The original gives 8/5. Our answer gives 8/5. They match. ✓

Watch out. Never add the bottoms. Once the bottoms match, they stay the same. If you subtract, the minus sign changes every term in the top that you are subtracting.

Key words. To solve an equation means to find the value of x that makes it true. Whatever you do to one side you must do to the other side, so that it stays balanced, like a pair of scales.

Method. Solve an equation containing fractions: (1) find the common denominator, (2) multiply every term by it to remove the fractions, (3) solve the simple equation that is left.

1. Write the equation. An equation says two things are equal. We want the value of x that makes both sides equal.
   x + 23 = x − 12
2. Find the common denominator of all the bottoms. Fractions are awkward to solve with, so we will remove them. To do that we multiply by something that every bottom divides into: the lowest common multiple of the bottoms.
   3, 2
   Common denominator = 6
3. Multiply every term, on both sides, by the common denominator. Doing the same thing to both sides keeps the equation balanced. Multiply every single term, including any plain numbers. This removes all the fractions.
   6 × x + 23 = 2(x + 2)   (the bottom 3 goes into 6 exactly 2 times, so the bottom disappears and the top is multiplied by 2)
   6 × x − 12 = 3(x − 1)   (the bottom 2 goes into 6 exactly 3 times, so the bottom disappears and the top is multiplied by 3)
4. Write the new equation. There are no fractions left. Check that every term from the original equation has been multiplied.
   2(x + 2) = 3(x − 1)
5. Expand the brackets. Multiply each term inside by the number or letter outside. A minus sign in front of a bracket changes the sign of every term inside it.
   Expand 2(x + 2):
   2 × x = 2x
   2 × 2 = 4
   so it is 2x + 4
   
   Expand 3(x − 1):
   3 × x = 3x
   3 × (−1) = −3
   so it is 3x − 3
6. Collect like terms on each side. Add or subtract the x terms together, and the plain numbers together, on each side separately.
   2x + 4 = 3x − 3
   2x + 4 = 3x − 3
7. Get the x terms on one side and the plain numbers on the other. Use inverse operations on both sides. The inverse of adding is subtracting and the inverse of subtracting is adding. Always do the same to both sides.
   Start: 2x + 4 = 3x − 3
   Subtract 2x from both sides (we move the smaller x term, to keep the numbers positive): 4 = x − 3
   Now add 3 to both sides: 7 = x
8. Read off the answer. x is already on its own.
   x = 7

Check. Put x = 7 back into the original equation. Left side: 9/3 = 3. Right side: 6/2 = 3. Both sides are equal, so the solution is correct. ✓

Watch out. Multiply every term, including the numbers that are not fractions. And a solution is only allowed if it does not make a bottom equal to 0.

Method. Turn each change into a multiplier and multiply the multipliers.

1. Find the multiplier for the increase. Increasing by 14 means 1 + 14 of the number.
   1 + 14 = 54
2. Find the multiplier for the decrease. Decreasing by 15 (of the new number) means 1 − 15 of it is left.
   1 − 15 = 45
3. Multiply the multipliers. The second change applies to the result of the first. Cancel the 5s and the 4s.
   54 × 45 = 2020 = 1
4. Conclude. A multiplier of 1 means no change.
   The result is the same as the original number.

Check. Try 40: increase by 1/4 gives 40 + 10 = 50. Decrease by 1/5 of 50 gives 50 − 10 = 40. We are back at 40. ✓

Watch out. This happens here because 5/4 and 4/5 are reciprocals. It is not true for other pairs, such as 1/5 up then 1/5 down.

Method. Write a recurring decimal as a fraction using algebra: line up the repeating parts, then subtract to cancel them.

1. Understand the aim. A recurring decimal never ends, so we cannot write it as a fraction by dividing. Instead we use algebra: we make two copies of the number whose repeating tails line up exactly, then subtract so the endless part disappears.
   0.35 = 0.3555…
2. Let x equal the recurring decimal. Using a letter lets us do algebra on the number.
   x = 0.35 = 0.3555…
3. Multiply by 10. This moves the non-repeating digit (3) to the left of the point so that the repeating digits start straight after the point.
   10x = 3.55…
4. Multiply by 100. This moves one more full block of repeating digits to the left of the point, so both lines have the same tail.
   100x = 35.55…
5. Subtract. Both lines have the same repeating tail after the point, so the tail cancels.
   100x − 10x = 35 − 3
   90x = 32
6. Solve for x. Divide both sides by the number in front of x.
   x = 3290
7. Simplify the fraction. The highest common factor of 32 and 90 is 2, so divide the top and the bottom by 2.
   3290 = 1645

Check. A different route: divide the top by the bottom. 16 ÷ 45 gives 0.35, which matches the original. ✓

Watch out. The number you multiply by (10, 100, 1000…) depends on how many digits repeat and how many do not. Both lines must have the same repeating tail.

Method. To divide by a fraction, turn the second fraction upside down and multiply: keep, change, flip.

1. Read the question and decide the plan. Dividing by a fraction is done by multiplying by its reciprocal. First look at the type of numbers we have.
   213 ÷ 134
2. Change the mixed numbers to improper fractions. A mixed number is a whole number plus a fraction. Multiply the whole number by the denominator, add the numerator, and keep the denominator. This lets us calculate with a single fraction.
   213 = (2 × 3 + 1) over 3 = 73
   134 = (1 × 4 + 3) over 4 = 74
3. Keep, change, flip. Dividing by a fraction is the same as multiplying by its reciprocal (the fraction turned upside down). For example, dividing by 1/2 asks "how many halves fit?", which is the same as multiplying by 2.
   73 ÷ 74 = 73 × 47
4. Cancel common factors. Divide a top and a bottom by a number they share, to keep the numbers small.
   73 × 47 = 13 × 41
5. Multiply the numerators, then the denominators. Now it is an ordinary multiplication: top × top and bottom × bottom.
   1 × 4 = 4
   3 × 1 = 3
   43
6. Write the improper fraction as a mixed number. The top is bigger than the bottom, so the answer is more than 1. Divide to find the whole number and the remainder.
   4 ÷ 3 = 1 remainder 1
   43 = 113

Check. Multiply back: 113 × 134 should give 213. 1.3333 × 1.75 = 2.3333 and 213 = 2.3333. ✓

Watch out. Only flip the second fraction (the one you are dividing by), never the first.

Method. Follow BIDMAS: Brackets, Indices (powers), Division and Multiplication, then Addition and Subtraction.

1. Decide the order of working. Brackets first. There are two brackets, so each must become a single fraction before we divide.
   (34 + 56) ÷ (1 − 23)
2. Work out the first bracket. The LCM of 4 and 6 is 12.
   34 + 56 = 912 + 1012 = 1912
3. Work out the second bracket. Write 1 as thirds: 1 = 3/3.
   1 − 23 = 33 − 23 = 13
4. Keep, change, flip. Dividing by a fraction is multiplying by its reciprocal.
   1912 ÷ 13 = 1912 × 31
5. Cancel and multiply. The 3 on top and the 12 on the bottom share a factor of 3.
   1912 × 31 = 194
6. Write as a mixed number. 19 ÷ 4 = 4 remainder 3.
   434

Check. As decimals: (0.75 + 0.833) ÷ (1 − 0.667) = 1.583 ÷ 0.333 = 4.75, which is 4 3/4. ✓

Watch out. Work out both brackets completely before dividing.

Method. Write a recurring decimal as a fraction using algebra: line up the repeating parts, then subtract to cancel them.

1. Understand the aim. A recurring decimal never ends, so we cannot write it as a fraction by dividing. Instead we use algebra: we make two copies of the number whose repeating tails line up exactly, then subtract so the endless part disappears.
   0.27 = 0.272727…
2. Let x equal the recurring decimal. Using a letter lets us do algebra on the number.
   x = 0.27 = 0.272727…
3. Multiply by 100. 2 digits repeat, so multiplying by 10² = 100 moves exactly one whole block of repeating digits to the left of the point. The tail after the point is then the same as before.
   100x = 27.2727…
4. Subtract the original x. Both numbers have the same infinite tail of digits after the point, so subtracting makes the tail disappear. Remember x means 1x.
   100x − x = 27.2727… − 0.2727…
   99x = 27
5. Solve for x. Divide both sides by the number in front of x.
   x = 2799
6. Simplify the fraction. The highest common factor of 27 and 99 is 9, so divide the top and the bottom by 9.
   2799 = 311

Check. A different route: divide the top by the bottom. 3 ÷ 11 gives 0.27, which matches the original. ✓

Watch out. The number you multiply by (10, 100, 1000…) depends on how many digits repeat and how many do not. Both lines must have the same repeating tail.

Method. Write a recurring decimal as a fraction using algebra: line up the repeating parts, then subtract to cancel them.

1. Plan the proof. In a “show that” question the answer is given, so every step must be shown. Start from one side and use correct algebra until you reach the other side.
   Let x = 0.58333…
2. Understand the aim. A recurring decimal never ends, so we cannot write it as a fraction by dividing. Instead we use algebra: we make two copies of the number whose repeating tails line up exactly, then subtract so the endless part disappears.
   0.583 = 0.58333…
3. Let x equal the recurring decimal. Using a letter lets us do algebra on the number.
   x = 0.583 = 0.58333…
4. Multiply by 100. This moves the non-repeating digits (58) to the left of the point so that the repeating digits start straight after the point.
   100x = 58.33…
5. Multiply by 1000. This moves one more full block of repeating digits to the left of the point, so both lines have the same tail.
   1000x = 583.33…
6. Subtract. Both lines have the same repeating tail after the point, so the tail cancels.
   1000x − 100x = 583 − 58
   900x = 525
7. Solve for x. Divide both sides by the number in front of x.
   x = 525900
8. Simplify the fraction. The highest common factor of 525 and 900 is 75, so divide the top and the bottom by 75.
   525900 = 712
9. State the result. The last line must match the result we were asked to show.
   0.58333… = 712 ✓

Check. A different route: divide the top by the bottom. 7 ÷ 12 gives 0.583, which matches the original. ✓

Watch out. The number you multiply by (10, 100, 1000…) depends on how many digits repeat and how many do not. Both lines must have the same repeating tail.

Method. Turn each reduction into a multiplier, multiply them, then work backwards from the final price.

1. Find the first multiplier. A reduction of 15 leaves 45.
   1 − 15 = 45
2. Find the second multiplier. A further reduction of 14 of the new price leaves 34.
   1 − 14 = 34
3. Find the overall multiplier. Multiply the multipliers. Cancel the 4s.
   45 × 34 = 35
4. Use the final price. The final price is 35 of the original. 3 parts are £90, so one fifth is £90 ÷ 3.
   1 fifth = £30
5. Find the original price. The original is 5 fifths.
   5 × £30 = £150

Check. Work forwards: 4/5 of £150 = £120, and 3/4 of £120 = £90. ✓

Watch out. Do not add the reductions 15 + 14. The second one applies to the new, smaller price.

Key words. A factor is something that is multiplied. For example 2 and 3 are factors of 6, and (x + 1) is a factor of 3(x + 1). Factorise means rewrite as a product. Cancel means divide the top and the bottom of a fraction by the same factor. An algebraic fraction is just a fraction with letters in it, and it follows exactly the same rules as number fractions.

Method. Simplify an algebraic fraction: (1) factorise the top, (2) factorise the bottom, (3) cancel the factors that match.

1. Write the question and say what is wanted. To simplify means: make the top and bottom as small as possible by cancelling every common factor. The answer must have no common factor left.
   x² + 5x + 6x² − 9
2. Factorise the top: x² + 5x + 6 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
   Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 5 and c = 6.
   We need two numbers that multiply to 6 (the last number) and add to 5 (the middle number). List the pairs that multiply to 6:
   1 × 6 = 6,   add: 1 + 6 = 7 ✗
   -1 × -6 = 6,   add: -1 + -6 = -7 ✗
   2 × 3 = 6,   add: 2 + 3 = 5 ✓ this pair works
   Put the pair into two brackets: (x + 2)(x + 3).
   Check by expanding: x² + 3x + 2x + 6 = x² + 5x + 6. ✓
   x² + 5x + 6 = (x + 2)(x + 3)
3. Factorise the bottom: x² − 9 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
   Spot the pattern. There are two terms, both are perfect squares, and there is a minus between them. This is a difference of two squares.
   The rule is: a² − b² = (a − b)(a + b).
   Find a and b. x² = (x)², so a = x. And 9 = 3², so b = 3.
   Put them into the rule: (x − 3)(x + 3).
   Check by expanding: (x − 3)(x + 3) = x² + 3x − 3x − 9 = x² − 9. ✓ (the middle terms cancel)
   x² − 9 = (x − 3)(x + 3)
4. Write the fraction again using the factorised top and bottom. Now every factor can be seen and compared.
   (x + 2)(x + 3)(x − 3)(x + 3)
5. Find the factors that are exactly the same on the top and the bottom, and cancel them. Dividing the top and the bottom by the same factor does not change the value of the fraction. A bracket on the top only cancels with a bracket on the bottom that is exactly the same.
   Matching factors: (x + 3).
   (x + 2)(x + 3)(x − 3)(x + 3)
   (the crossed-out factors cancel)
6. Write what is left. This is the answer. Multiply together whatever has not been crossed out.
   (x + 2)(x − 3)

Check. Choose a number for x that does not make any bottom zero. Try x = 2. The original: top = 20, bottom = -5, so the fraction = -4. Our answer at x = 2 = -4. They match. ✓

Watch out. You can only cancel factors (things multiplied), never terms (things added). In (x + 6)/x you cannot cancel the x, because the x is added to 6 on top. That is why we always factorise first.

Key words. A factor is something that is multiplied. For example 2 and 3 are factors of 6, and (x + 1) is a factor of 3(x + 1). Factorise means rewrite as a product. Cancel means divide the top and the bottom of a fraction by the same factor. An algebraic fraction is just a fraction with letters in it, and it follows exactly the same rules as number fractions.

Method. Multiply or divide algebraic fractions: (1) flip the fraction you are dividing by, (2) factorise every top and bottom, (3) write as one fraction, (4) cancel matching factors.

1. Keep, change, flip. Dividing by a fraction is the same as multiplying by its reciprocal, which is the fraction turned upside down. Here is why it works with numbers: 1/2 ÷ 1/4 asks "how many quarters fit into a half?" The answer is 2, and 1/2 × 4/1 = 2. So: keep the first fraction, change ÷ to ×, and flip the second fraction.
   x² − 4x + 3 ÷ x − 2x² + 3x = x² − 4x + 3 × x² + 3xx − 2
2. Factorise the first top: x² − 4 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
   Spot the pattern. There are two terms, both are perfect squares, and there is a minus between them. This is a difference of two squares.
   The rule is: a² − b² = (a − b)(a + b).
   Find a and b. x² = (x)², so a = x. And 4 = 2², so b = 2.
   Put them into the rule: (x − 2)(x + 2).
   Check by expanding: (x − 2)(x + 2) = x² + 2x − 2x − 4 = x² − 4. ✓ (the middle terms cancel)
   x² − 4 = (x − 2)(x + 2)
3. Factorise the first bottom: x + 3 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
   x + 3 is a single bracket already. It cannot be factorised any further.
   x + 3 = (x + 3)
4. Factorise the second top: x² + 3x Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
   Look for a common factor. A common factor is something that divides into every term.
   The terms are x², 3x. The biggest number and the highest power of x that divide into all of them is x.
   Take it outside a bracket, and inside the bracket put each term divided by it:
   x² ÷ x = x
   3x ÷ x = 3
   So x² + 3x = x(x + 3).
   Check by expanding: x × (x + 3) = x × x, x × 3 → x² + 3x. ✓
   x² + 3x = x(x + 3)
5. Factorise the second bottom: x − 2 Factorising means writing it as a product (things multiplied together). We need the factors to be visible so that matching ones can cancel.
   x − 2 is a single bracket already. It cannot be factorised any further.
   x − 2 = (x − 2)
6. Write the multiplication again using all the factorised pieces. This lets us see every factor at once.
   (x − 2)(x + 2)(x + 3) × x(x + 3)(x − 2)
7. Write it as a single fraction. To multiply fractions, multiply the tops together and multiply the bottoms together. Do not multiply out the brackets. Keep the factors so that we can cancel them.
   (x − 2)(x + 2) × x(x + 3)(x + 3) × (x − 2)
8. Cancel every factor that appears on both the top and the bottom. A factor on the top cancels with an identical factor on the bottom because we are dividing the top and the bottom by it. The brackets must be exactly the same.
   Matching factors: (x − 2), (x + 3).
   (x − 2)x(x + 2)(x + 3)(x − 2)(x + 3)
   (the crossed-out factors cancel)
9. Write what is left. This is the answer. Multiply together the factors that have not been crossed out.
   x(x + 2)

Check. Try x = 3 (it does not make any bottom zero). The original expression gives 15. Our answer gives 15. They match. ✓

Watch out. Only flip the second fraction, never the first. Only cancel matching factors, never terms that are added.

Key words. The denominator is the bottom of a fraction and the numerator is the top. A common denominator is a bottom that all the fractions share. This works exactly like 1/2 + 1/3 = 3/6 + 2/6 = 5/6, except that the bottoms contain letters.

Method. Add or subtract algebraic fractions: (1) common denominator, (2) rewrite each fraction, (3) combine the tops, (4) expand and collect, (5) simplify if possible.

1. Write the question. We are adding or subtracting algebraic fractions. The answer should be one single fraction.
   2x + 1 + 3x − 2
2. Find a common denominator. We can only add or subtract pieces that are the same size, so the bottoms must match. The common denominator is the smallest expression that every bottom divides into. Method: factorise each bottom, then take every different factor, using the highest number of copies of it that appear in any one bottom.
   x + 1 = (x + 1)
   x − 2 = (x − 2)
   Common denominator = (x − 2)(x + 1)
3. Rewrite every fraction with the common denominator. Multiplying the top and the bottom by the same thing is like multiplying by 1, so the value of the fraction does not change.
   2x + 1: to turn (x + 1) into (x − 2)(x + 1) we multiply the bottom by (x − 2). We must do the same to the top, so the value does not change:
   2(x − 2)(x − 2)(x + 1)
   3x − 2: to turn (x − 2) into (x − 2)(x + 1) we multiply the bottom by (x + 1). We must do the same to the top, so the value does not change:
   3(x + 1)(x − 2)(x + 1)
4. Write one fraction: combine the tops over the common denominator. Now the pieces are the same size, so we add or subtract the tops and keep the bottom.
   2(x − 2) + 3(x + 1)(x − 2)(x + 1)
5. Expand each product on the top. Multiply every term in one bracket by every term in the other.
   Expand 2(x − 2):
   2 × x = 2x
   2 × (−2) = −4
   so 2(x − 2) = 2x − 4
   
   Expand 3(x + 1):
   3 × x = 3x
   3 × 1 = 3
   so 3(x + 1) = 3x + 3
6. Write the top without brackets. There is no minus sign in front of a bracket, so the brackets can simply be removed and the terms written in a row.
   Top = 2x − 4 + 3x + 3
7. Collect like terms. Like terms have the same letter part. Add or subtract all the x terms together, and all the plain numbers together.
   Top = 5x − 1
8. Write the answer. Check if it can be simplified. Leave the bottom factorised. See if the top shares a factor with the bottom: if not, we are done.
   5x − 1(x − 2)(x + 1)
   The top does not share a factor with the bottom, so this is the simplest form.

Check. Try x = 3 (it does not make any bottom zero). The original gives 7/2. Our answer gives 7/2. They match. ✓

Watch out. Never add the bottoms. Once the bottoms match, they stay the same. If you subtract, the minus sign changes every term in the top that you are subtracting.

Key words. To solve an equation means to find the value of x that makes it true. Whatever you do to one side you must do to the other side, so that it stays balanced, like a pair of scales.

Method. Solve an equation containing fractions: (1) find the common denominator, (2) multiply every term by it to remove the fractions, (3) solve the simple equation that is left.

1. Write the equation. An equation says two things are equal. We want the value of x that makes both sides equal.
   2x + 13 − x − 12 = 1
2. Find the common denominator of all the bottoms. Fractions are awkward to solve with, so we will remove them. To do that we multiply by something that every bottom divides into: the lowest common multiple of the bottoms.
   3, 2
   Common denominator = 6
3. Multiply every term, on both sides, by the common denominator. Doing the same thing to both sides keeps the equation balanced. Multiply every single term, including any plain numbers. This removes all the fractions.
   6 × 2x + 13 = 2(2x + 1)   (the bottom 3 goes into 6 exactly 2 times, so the bottom disappears and the top is multiplied by 2)
   6 × x − 12 = 3(x − 1)   (the bottom 2 goes into 6 exactly 3 times, so the bottom disappears and the top is multiplied by 3)
   6 × 1 = 6   (a plain number or term with no bottom is simply multiplied by 6)
4. Write the new equation. There are no fractions left. Check that every term from the original equation has been multiplied.
   2(2x + 1) − 3(x − 1) = 6
5. Expand the brackets. Multiply each term inside by the number or letter outside. A minus sign in front of a bracket changes the sign of every term inside it.
   Expand 2(2x + 1):
   2 × 2x = 4x
   2 × 1 = 2
   so it is 4x + 2
   
   Expand 3(x − 1):
   3 × x = 3x
   3 × (−1) = −3
   so it is 3x − 3
6. Collect like terms on each side. Add or subtract the x terms together, and the plain numbers together, on each side separately.
   4x + 2 − 3x + 3 = 6
   x + 5 = 6
7. Get the x terms on one side and the plain numbers on the other. Use inverse operations on both sides. The inverse of adding is subtracting and the inverse of subtracting is adding. Always do the same to both sides.
   Start: x + 5 = 6
   Now get the numbers on the other side. Subtract 5 from both sides: x = 1
8. Read off the answer. x is already on its own.
   x = 1

Check. Put x = 1 back into the original equation. Left side: 3/3 − 0/2 = 1. Right side: 1 = 1. Both sides are equal, so the solution is correct. ✓

Watch out. Multiply every term, including the numbers that are not fractions. And a solution is only allowed if it does not make a bottom equal to 0.

Key words. To solve an equation means to find the value of x that makes it true. Whatever you do to one side you must do to the other side, so that it stays balanced, like a pair of scales.

Method. Solve an equation containing fractions: (1) find the common denominator, (2) multiply every term by it to remove the fractions, (3) solve the simple equation that is left.

1. Write the equation. An equation says two things are equal. We want the value of x that makes both sides equal.
   5x + 1 = 2x − 2
2. Find the common denominator of all the bottoms. Fractions are awkward to solve with, so we will remove them. To do that we multiply by something that every bottom divides into: the lowest common multiple of the bottoms.
   x + 1, x − 2
   Common denominator = (x − 2)(x + 1)
3. Multiply every term, on both sides, by the common denominator. Doing the same thing to both sides keeps the equation balanced. Multiply every single term, including any plain numbers. This removes all the fractions.
   (x² − x − 2) × 5x + 1 = 5(x − 2)   (the bottom x + 1 goes into (x − 2)(x + 1) exactly (x − 2) times, so the bottom disappears and the top is multiplied by (x − 2))
   (x² − x − 2) × 2x − 2 = 2(x + 1)   (the bottom x − 2 goes into (x − 2)(x + 1) exactly (x + 1) times, so the bottom disappears and the top is multiplied by (x + 1))
4. Write the new equation. There are no fractions left. Check that every term from the original equation has been multiplied.
   5(x − 2) = 2(x + 1)
5. Expand the brackets. Multiply each term inside by the number or letter outside. A minus sign in front of a bracket changes the sign of every term inside it.
   Expand 5(x − 2):
   5 × x = 5x
   5 × (−2) = −10
   so it is 5x − 10
   
   Expand 2(x + 1):
   2 × x = 2x
   2 × 1 = 2
   so it is 2x + 2
6. Collect like terms on each side. Add or subtract the x terms together, and the plain numbers together, on each side separately.
   5x − 10 = 2x + 2
   5x − 10 = 2x + 2
7. Get the x terms on one side and the plain numbers on the other. Use inverse operations on both sides. The inverse of adding is subtracting and the inverse of subtracting is adding. Always do the same to both sides.
   Start: 5x − 10 = 2x + 2
   The x terms should be on one side. Subtract 2x from both sides: 3x − 10 = 2
   Now get the numbers on the other side. Add 10 to both sides: 3x = 12
8. Divide both sides by the number in front of x (which is 3). 3x means 3 multiplied by x. The inverse of multiplying by 3 is dividing by 3. This leaves x on its own.
   x = 12 ÷ 3 = 4

Check. Put x = 4 back into the original equation. Left side: 5/5 = 1. Right side: 2/2 = 1. Both sides are equal, so the solution is correct. ✓

Watch out. Multiply every term, including the numbers that are not fractions. And a solution is only allowed if it does not make a bottom equal to 0.

Method. Use a common denominator to subtract the fractions, expand and collect the top, and compare with the result.

1. Plan the proof. In a “show that” or “prove” question every step must be shown and the last line must be the result we were given. Using the letter n means the working is true for every value of n, not just one example.
   Start from the left side: 1n − 1n + 1
2. Find a common denominator. We can only subtract fractions with the same bottom. The bottoms n and n + 1 have no common factor (they are consecutive numbers), so the lowest common denominator is their product.
   Common denominator = n(n + 1)
3. Rewrite the first fraction with denominator n(n + 1). Multiply the top and the bottom by (n + 1), the factor that is missing from its bottom. This is multiplying by 1, so the value does not change.
   1n = 1 × (n + 1)n(n + 1)
4. Rewrite the second fraction with denominator n(n + 1). Multiply the top and the bottom by n.
   1n + 1 = 1 × nn(n + 1)
5. Combine the tops over the common denominator. Now the pieces are the same size, so we subtract the tops and keep the bottom. The bracket around the second top is needed because we subtract all of it.
   1 × (n + 1) − (1 × n)n(n + 1)
6. Expand each product on the top. Multiply every term in one bracket by every term in the other bracket.
   Expand 1 × (n + 1):
   1 × n = n
   1 × 1 = 1
   so 1 × (n + 1) = n + 1
   
   1 × n = n
7. Remove the brackets. A minus sign in front of a bracket changes the sign of every term inside it.
   Top = (n + 1) − n
   Top = n + 1 − n
8. Collect like terms. Add or subtract the terms that have the same letter part, and the plain numbers together.
   Top = 1
9. Write the result. Put the simplified top over the common denominator. The top has nothing in common with the bottom, so nothing cancels.
   1n(n + 1)
10. State the result. The last line matches the result we were asked to show.
    1n − 1n + 1 = 1n(n + 1) ✓

Check. Try n = 2. The original gives 1/6. Our result gives 1/6. They are equal, so the algebra is right. (One example does not prove it, but it catches slips.) ✓

Watch out. Never add or subtract the bottoms. And when you subtract a fraction, the minus sign applies to every term in its top.

Method. For work-rate problems, find the fraction of the tank each pipe fills in one hour, add them, then turn the combined rate into a time.

1. Find how much pipe A fills in one hour. A fills the tank in 6 hours, so in 1 hour it fills one sixth.
   A: 16 per hour
2. Find how much pipe B fills in one hour. B fills the tank in 3 hours, so in 1 hour it fills one third.
   B: 13 per hour
3. Add the rates. Working together they fill both amounts each hour. The LCM of 6 and 3 is 6.
   16 + 26 = 36 = 12 per hour
4. Find the time. If half the tank is filled each hour, the whole tank takes 1 ÷ 12 hours.
   1 ÷ 12 = 2 hours

Check. In 2 hours A fills 26 = 13 and B fills 23. 13 + 23 = 1 whole tank. ✓

Watch out. Add the rates (fractions per hour), not the times.

Method. Simplify the fraction, write its denominator as a product of primes, and see whether any prime other than 2 or 5 appears.

1. Say what the test is. A fraction in its simplest form gives a terminating decimal only if its denominator has no prime factors other than 2 and 5. This is because 10 = 2 × 5, so only denominators made of 2s and 5s can be turned into tenths, hundredths and so on.
   1180
2. Check the fraction is in its simplest form. Common factors could change the denominator, so we check first.
   11 and 80 share no factor except 1, so 1180 is already simplest.
3. Break the denominator into prime factors. A prime factor is a prime number that divides into it. We divide by the smallest prime each time.
   80 = 2⁴ × 5
   80 ÷ 2 = 40
   40 ÷ 2 = 20
   20 ÷ 2 = 10
   10 ÷ 2 = 5
   5 ÷ 5 = 1
4. Look for primes other than 2 and 5. If there is any other prime factor, the decimal can never end.
   Only 2s and 5s appear, so the decimal terminates.
5. Write the conclusion. Answer the question that was asked.
   1180 gives a terminating decimal.

Check. A different route: divide. 11 ÷ 80 = 0.1375 (it ends). ✓

Watch out. Always simplify first. For example 6/15 = 2/5 terminates even though 15 has a factor of 3.

Method. Find the width from the area and the length, then use the perimeter to form an equation for x.

1. Find the width. Area = length × width, so width = area ÷ length.
   Width = x² + 5x + 6x + 3
2. Factorise the top. Find two numbers that multiply to 6 and add to 5: they are 2 and 3.
   x² + 5x + 6 = (x + 2)(x + 3)
3. Cancel the common bracket. (x + 3) is on the top and the bottom.
   Width = (x + 2) cm
4. Write the perimeter. Perimeter = 2 × (length + width).
   2((x + 3) + (x + 2)) = 2(2x + 5)
5. Write the equation. The perimeter is 20 cm.
   2(2x + 5) = 20
6. Solve. Divide both sides by 2, subtract 5, then divide by 2.
   2x + 5 = 10
   2x = 5
   x = 2.5

Check. Put x = 2.5 back: length = 5.5 cm, width = 4.5 cm. Area = 5.5 × 4.5 = 24.75 and x² + 5x + 6 = 6.25 + 12.5 + 6 = 24.75. Perimeter = 2(5.5 + 4.5) = 20. ✓

Watch out. Find the width first. The perimeter equation needs both the length and the width.

Method. Use a common denominator to subtract the fractions, expand and collect the top, and compare with the result.

1. Plan the proof. In a “show that” or “prove” question every step must be shown and the last line must be the result we were given. Using the letter n means the working is true for every value of n, not just one example.
   Start with the difference: nn + 1 − n − 1n
2. Find a common denominator. We can only subtract fractions with the same bottom. The bottoms n + 1 and n have no common factor (they are consecutive numbers), so the lowest common denominator is their product.
   Common denominator = n(n + 1)
3. Rewrite the first fraction with denominator n(n + 1). Multiply the top and the bottom by n, the factor that is missing from its bottom. This is multiplying by 1, so the value does not change.
   nn + 1 = n × nn(n + 1)
4. Rewrite the second fraction with denominator n(n + 1). Multiply the top and the bottom by (n + 1).
   n − 1n = (n − 1) × (n + 1)n(n + 1)
5. Combine the tops over the common denominator. Now the pieces are the same size, so we subtract the tops and keep the bottom. The bracket around the second top is needed because we subtract all of it.
   n × n − ((n − 1) × (n + 1))n(n + 1)
6. Expand each product on the top. Multiply every term in one bracket by every term in the other bracket.
   n × n = n²
   
   Expand (n − 1) × (n + 1):
   First: n × n = n²
   Outer: n × 1 = n
   Inner: (−1) × n = −n
   Last: (−1) × 1 = −1
   so (n − 1) × (n + 1) = n² − 1
7. Remove the brackets. A minus sign in front of a bracket changes the sign of every term inside it.
   Top = n² − (n² − 1)
   Top = n² − n² + 1
8. Collect like terms. Add or subtract the terms that have the same letter part, and the plain numbers together.
   Top = 1
9. Write the result. Put the simplified top over the common denominator. The top has nothing in common with the bottom, so nothing cancels.
   1n(n + 1)
10. Decide the sign. The difference is 1 divided by n(n + 1). When n is a positive integer, both n and n + 1 are positive, so the bottom is positive, and the top is 1. A positive number divided by a positive number is positive.
    1n(n + 1) > 0 for every positive integer n
11. State the result. The last line matches the result we were asked to show.
    nn + 1 > n − 1n for every positive integer n ✓

Check. Try n = 2. The original gives 1/6. Our result gives 1/6. They are equal, so the algebra is right. (One example does not prove it, but it catches slips.) ✓

Watch out. Never add or subtract the bottoms. And when you subtract a fraction, the minus sign applies to every term in its top.